*Update: Thanks to Anschel for noting that I messed up the statement of the last exercise. It’s fixed now. Thanks to Justin for noting that I messed up a square root. Pythagorean theorem is hard, yo.*

About a year and a half ago I explained what hyperbolic space is, specifically by contrasting it with Euclidean space and spherical space. We’ve also run into hyperbolic groups a few times, which are groups whose Cayley graphs are somehow like hyperbolic space. More precisely, a group is hyperbolic if, whenever you have a Cayley graph of that group, triangles are thin, which means the third side of any triangle is contained in a neighborhood of the other two sides. It’s important that the same works for **every **triangle in the space.

Note that triangles in Euclidean space are way totally far from being hyperbolic. For any big number *n*, you can make a triangle so that the third side is **not **contained in an *n-*neighborhood of the other two sides: just take a *2n *horizontal segment and a *2n *vertical segment to make an isoceles right triangle. If *n *is bigger than 2, then the midpoint of the hypotenuse is farther than *n *away from the other two sides. As usual, this long paragraph could be better done in a picture.

Anyways, I just put in that definition because it’s the first thing you’ll hear or see in a colloquium talk that involves the word “hyperbolic.” Let’s play with the upper half plane model of hyperbolic space! Here’s a repeat picture from that October 2014 post (wow that’s when baby was born! He’s walking around and getting into trouble now, btw.).

The graph paper lines in this picture are misleading; they contrast hyperbolic geodesics with Euclidean ones. So the gray lines are Euclidean geodesics, and the colored ones are hyperbolic. All geodesics in this model are either straight lines perpendicular to the horizontal axis, or semicircles perpendicular to the horizontal axis. All of the horizontal axis and everything that the straight up and down geodesics end at (sort of like a horizontal axis infinitely far away) represent infinity.

I’ll write down the metric in case you were wondering, but we won’t need it for what we’ll be doing: [I took this formulation from wikipedia]. What this says is that the hyperbolic metric is a lot like the Euclidean one, except that the higher up you go on the y-axis, the less distance is covered (because of that 1/y factor). More precisely, if you’re just looking at the straight line geodesics, the distance between two points at heights *a<b *is .

The other fact we might want to know is that things that look like Euclidean dilations (stretching something like your pupil dilates from looking in a bright light to a dark room) are *isometries *in this model. You can see that in the picture above: the lines look like they’re stretching longer and longer in the Euclidean metric, but they’re actually all the same length. Speaking of isometries, if you have any two geodesics (like a vertical line and a big old semi-circle somewhere else), you can find an isometry that sends one to the other.

First question: what do circles look like? Whenever you have a metric space, it’s nice to know what neighborhoods look like, and the first thing you might want to consider are neighborhoods of points. Turns out circles in this model look like circles in Euclidean space, but the centers aren’t where you think they are. For instance, here’s a picture of circles with radius ln(2), which we saw in the straight lines above.

Notice that the centers of these circles hang a lot lower in the circle than they do in the Euclidean metric! Isn’t playtime fun?!

Generally when I play with math I throw out a lot of garbage ideas, and then eventually one of them is somewhat right. Other people apparently think for awhile before they put out an idea. Anyways, here are some sketches of what I thought a 2-neighborhood of a vertical line might look like:

Maybe you looked at these and were like “Yen that is nonsense what were you thinking?!” Maybe you are my advisor or a practiced mathematician. Let’s go through the nonsense-ness of each of these pictures:

The rightmost picture is a 2-neighborhood of the vertical line in Euclidean space. We know hyperbolic space is pretty drastically different from Euclidean space, so we wouldn’t expect the neighborhoods to be so similar. The middle and left pictures have similar shapes but different curviness, and yes we’d expect a hyperbolic neighborhood to look different so those are guesses based in some more intution. However, let’s try to figure out the actual size of a neighborhood of a vertical line. We can use our previous pictures, and switch to a ln(2) neighborhood.

Here I moved all our ln(2) circles so that their centers laid on the same line. A neighborhood of a line is just the union of the neighborhoods of all of the points on that line, so if we just keep making ln(2) circles along the line we’ll end up with a neighborhood of the whole line. So you can see that our actual neighborhood ended up being upside down from my middle picture above. If this explanation didn’t make sense, here’s [half] a 2-neighborhood of a Euclidean line:

Actually using Euclidean intuitions and then mixing them up a bit is a great way to play with the hyperbolic plane. This next exercise was an actual exercise in the book but it is just **so crazy **I have to share it with you. It’s just dramatically different from Euclidean space, just like the triangles were.

If you have a circle in the hyperbolic plane and * project *it to a geodesic segment that it doesn’t intersect (which means for any point on the circle, you find the closest point to it on the geodesic and draw a dot on the geodesic there), the projection is shorter than . Here’s the picture in Euclidean space where this makes no sense:

And here’s a picture in hyperbolic space that might make you think this could possibly just maybe be true. Any circle will eventually fit inside a big huge circle that looks like the blue one in the picture, so its projection would be shorter than the projection of the blue one. That means you only have to worry about big huge circles in that particular position. And by “big huge,” I mean “of (Euclidean) radius *n*“.

Remember, if we’re just looking at vertical lines, we know how to measure distance: it’s . So if you can show that the small orange circle hits the vertical line at and the big orange circle hits it at , you’ll have proved the contraction property. Try using Euclidean geometry, and think about how we did the triangles case.

That was fun for me I hope it was fun for you!

Trying to understand the argument at the end: am I right that this only works if we have some isometries that can transform any line into one of these “straight to infinity” kinds? What are those isometries?

Also, what if the geodesic passes through the circle?

Yes, great, I flew the fact that you need to map your geodesic to a vertical line under the radar. I actually left out the complex plane entirely in this post, but this upper half plane model of hyperbolic space means upper half of the complex plane, not Euclidean, if you’re actually going to make calculations. Fact: whenever you have two geodesics in the hyperbolic plane, you can build an isometry from one to the other. General isometries of the plane are given by SL(2,R). If you have a matrix (a b \\ c d), then the action of that matrix sends the point z to (az+b)/(cz+d). So the set up of this problem implies that step 1 is find an isometry to map your geodesic of choice to the imaginary axis. Step 2 is find an N large enough so that the image under this isometry of your ball of choice fits inside it.

Thanks for the disjointedness catch I missed that! In fact it’s definitely not true if the circle and geodesic are not disjoint (just make a bigger and bigger circle over the geodesic). I’ll fix that in the post.

On Wed, Feb 3, 2016 at 3:43 PM, Baking and Math wrote:

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